If the time of flight of a bullet over a horizontal range $R$ is $T$ , then inclination of the direction of projection with the horizontal is

{tan}^{- 1} [\frac{g T^{2}}{2 R}]

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{tan}^{- 1} [\frac{2 R^{2}}{g T}]

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{tan}^{- 1} [\frac{2 R}{g^{2} π}]

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{tan}^{- 1} [\frac{2 R}{g T}]

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Solution

The correct option is A ${tan}^{- 1} [\frac{g T^{2}}{2 R}]$
Time of flight ( $T$ ) $= \frac{2 v sin θ}{g}$
$T = \frac{2 v sin θ}{g^{2}}$
$\begin{matrix} T^{2} = \frac{4 v^{2} {sin}^{2} θ}{g^{2}} \\ (1) \end{matrix}$
We know that,
Range, $\begin{matrix} R = \frac{v^{2} (2 sin θ cos θ)}{g} \\ (2) \end{matrix}$
Dividing (1) from (2), we get
$\frac{T^{2}}{R} = \frac{2}{g} tan θ$
$tan θ = \frac{g T^{2}}{2 R}$
$θ = {tan}^{- 1} [\frac{g T^{2}}{2 R}]$

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